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                        <h1 class="single-title flipInX">区域检测——Harris角点</h1><div class="post-meta summary-post-meta"><span class="post-category meta-item">
                                <a href="/categories/%E8%AE%A1%E7%AE%97%E6%9C%BA%E8%A7%86%E8%A7%89/"><span class="svg-icon icon-folder"></span>计算机视觉</a>
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                                <span class="svg-icon icon-clock"></span><time class="timeago" datetime="2020-07-17">2020-07-17</time>
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                        <div class="details-summary toc-title">
                            <span>目录</span>
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                        <div class="details-content toc-content" id="toc-content-static"><nav id="TableOfContents">
  <ul>
    <li><a href="#解决思路">解决思路</a></li>
    <li><a href="#特征点corner">特征点——Corner</a>
      <ul>
        <li><a href="#好的特征点是什么样的">好的特征点是什么样的？</a></li>
        <li><a href="#什么样的点满足条件">什么样的点满足条件</a></li>
        <li><a href="#basic-idea">Basic Idea</a></li>
      </ul>
    </li>
    <li><a href="#数学描述">数学描述</a>
      <ul>
        <li><a href="#矩阵m">矩阵$M$</a></li>
      </ul>
    </li>
    <li><a href="#harris检测器">Harris检测器</a></li>
  </ul>
</nav></div>
                    </div><p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717204500384.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717204500384.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p>对于图像处理时经常需要提取特征点分析图片结构，将照片进行拼接，实现全景拍摄，那么在照片特征点提取时所采用的具体算法是什么呢？</p>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717112622949.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717112622949.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<h2 id="解决思路" class="headerLink"><a href="#%e8%a7%a3%e5%86%b3%e6%80%9d%e8%b7%af" class="header-mark"></a>解决思路</h2><ol>
<li>提取特征点</li>
<li>匹配特征点</li>
<li>使用<strong>RANSAC</strong>方法将两张图片的对应的特征点转换的方式拟合出来，在对图片采用相同的转换方式进行转换，在进行拼接</li>
</ol>
<h2 id="特征点corner" class="headerLink"><a href="#%e7%89%b9%e5%be%81%e7%82%b9corner" class="header-mark"></a>特征点——Corner</h2><h3 id="好的特征点是什么样的" class="headerLink"><a href="#%e5%a5%bd%e7%9a%84%e7%89%b9%e5%be%81%e7%82%b9%e6%98%af%e4%bb%80%e4%b9%88%e6%a0%b7%e7%9a%84" class="header-mark"></a>好的特征点是什么样的？</h3><ol>
<li>可重复性：在一张图可以被观测到的，在其他同场景的图也可以被观测到</li>
<li>显著性：检测的特征点需要是在某一类图像中“独有的”，尽量剔除“普遍性”的点，目的是为了将不同类的图区分开</li>
<li>简洁和高效：尽可能的减少计算量，提高计算效率</li>
<li>局部性：匹配特征时要匹配特征点之间的相对关系，通过局部特征相对位置来判断是否为同一张图，来拟合转动镜头角度，图像位置</li>
</ol>
<h3 id="什么样的点满足条件" class="headerLink"><a href="#%e4%bb%80%e4%b9%88%e6%a0%b7%e7%9a%84%e7%82%b9%e6%bb%a1%e8%b6%b3%e6%9d%a1%e4%bb%b6" class="header-mark"></a>什么样的点满足条件</h3><p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717114024178.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717114024178.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p>通过观察图片的特征，发现存在“<strong>角</strong>”的地方承载着更多的信息，角点是梯度在两个或以上方向上有变化的点。</p>
<h3 id="basic-idea" class="headerLink"><a href="#basic-idea" class="header-mark"></a>Basic Idea</h3><ol>
<li>使用一个较小的窗口在图像上延各个方向滑动</li>
<li>不同的变化趋势显示了不同的特征</li>
<li>图像内部所在的窗口延各个方向都没有变化；边缘所在的窗口延边缘方向无变化；角点所在窗口会在各个方向上都有显著的变化</li>
</ol>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717140905906.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717140905906.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<h2 id="数学描述" class="headerLink"><a href="#%e6%95%b0%e5%ad%a6%e6%8f%8f%e8%bf%b0" class="header-mark"></a>数学描述</h2><p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717142205136.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717142205136.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<ul>
<li>$u$和$v$是平移量</li>
<li>求平移后的窗口与平移前的窗口的对应位置差的平方，再累加</li>
<li>乘上窗口权重，考虑每个点对窗口影响的不同程度，例如第二种的高斯函数权重，就是考虑中间的点的差值在整个窗口的影响度更大</li>
<li><strong>二阶泰勒展开</strong>：为了能够直接观察到$E(u,v)$与$[u,v]$之间的联系
<blockquote>
<p>取$E(u,v)$在$(0,0)$的二阶展开作为近似解</p>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/2020071717034199.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/2020071717034199.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p>计算化简泰勒展开式</p>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717173020112.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717173020112.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p>其中$I_x,I_y$分别表示点$(x,y)$在$x$方向,$y$方向的偏导，$M$是由一个二阶矩矩阵加权求和得到</p>
</blockquote>
</li>
</ul>
<h3 id="矩阵m" class="headerLink"><a href="#%e7%9f%a9%e9%98%b5m" class="header-mark"></a>矩阵$M$</h3><p>类比方程$y=ax+b$决定方程特性的是$a,b$。则决定$E(u,v)$特性的是$M$，分析矩阵$M$就可以得到$E(u,v)$的特性</p>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717182032753.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717182032753.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p>函数图像延竖直方向截取为一个椭圆，当梯度为零时，截面为圆，此时窗口位于图像内部；当延某一方向梯度为零时，界面为一个“正椭圆”，此时窗口位于边；当窗口位于角时，界面椭圆的形状反映了当前窗口下角的特性</p>
<ul>
<li>
<p>正交矩阵$R$：使所截取的椭圆旋转$R$角度，变为一个“正椭圆”</p>
<blockquote>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/2020071719160637.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/2020071719160637.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<ol>
<li>当$I_x,I_y$任意一个趋于0，用$\lambda$表示，任意一个$\lambda$趋于0，都表示这个点不是<strong>角点</strong></li>
<li>椭圆的半轴长度反应的是梯度变化的快慢，越长则梯度变化越快（将$E(u,v)$展开就可以得到椭圆的半轴表示为$\lambda^{-\frac12}$）</li>
</ol>
</blockquote>
</li>
<li>
<p>可视化</p>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717192818265.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717192818265.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
</li>
<li>
<p>特征值简化——角点响应函数R</p>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717195427923.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717195427923.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p>将$\lambda_1,\lambda_2$特征转化给$R$，<strong>最后判断只需判断$R$就可以确定是否为角点</strong></p>
</li>
</ul>
<h2 id="harris检测器" class="headerLink"><a href="#harris%e6%a3%80%e6%b5%8b%e5%99%a8" class="header-mark"></a>Harris检测器</h2><ol>
<li>计算每个像素处的高斯导数</li>
<li>计算每个像素周围的高斯窗口中的二阶矩矩阵M</li>
<li>计算角点响应函数R</li>
<li>设置门限R</li>
<li>寻找响应函数的局部最大值(非最大抑制)</li>
</ol>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717201858186.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717201858186.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p><strong>Harris的特性</strong></p>
<ol>
<li>当光线强度，明暗改变时，只是改变了部分角点的值，还有大部分的点可以用于检测，可以进行检测，</li>
<li>当改变位置，角度时，没有改变相对位置，可以检测</li>
<li><strong>当改变窗口大小时</strong>，大窗口下是角点，而小窗口下是线或者边缘，<strong>无法检测</strong></li>
</ol>
<p>




<figure class="render-image"><a target="_blank" href="https://img-blog.csdnimg.cn/20200717203432876.png" title=" " >
        <img loading="lazy" decoding="async"
             class="render-image"
             src="https://img-blog.csdnimg.cn/20200717203432876.png"
            alt=" "
        />
    </a><figcaption class="image-caption"> </figcaption>
</figure></p>
<p><strong>学习资源：<a href="https://www.bilibili.com/video/BV1nz4y197Qv" target="_blank" rel="noopener noreffer">北京邮电大学计算机视觉——鲁鹏</a></strong></p>
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